Recognitions:

Laplacian of a 2-form

Hi,

According to eg Nakahara's conventions the Laplacian on a form K is given by

$$\Delta K = (dd^{\dagger} + d^{\dagger}d)K$$

In my case K is a two form living in R^3. I've calculated the Laplacian and arrive at

$$\Delta K = \Bigl( \frac{1}{3!}\epsilon^{klm}\epsilon^n_{\ ij}\partial_k \partial_n K_{lm} - \frac{1}{4}\partial_{i}\partial^k K_{jk} \Bigr) dx^i \wedge dx^j$$

However, the answer seems a little odd to me. Does anyone have a reference where Laplacians of two forms are evaluated, or some comment? Thanks in forward! :)
 PhysOrg.com science news on PhysOrg.com >> King Richard III found in 'untidy lozenge-shaped grave'>> Google Drive sports new view and scan enhancements>> Researcher admits mistakes in stem cell study
 Recognitions: Science Advisor And another question: is on a compact manifold a harmonic form neccesarily constant, or is this only true for 0-forms (functions)?
 How would you go about defining a constant form? The obvious answer to this would be that you can't define it in a coordinate-invariant way. The generalization of the idea of constant functions to forms is with the definition: a constant function is d and d* closed.

Laplacian of a 2-form

For your first question, if the codifferential is defined with a sign (-1)^n(p+1), it can be shown that in euclidean space in general laplace beltrami of a form is the same as the usual old laplacian of every function factor (times the wedged covectors). It is a lengthy calculation and took me some hours, but it works ;)
 Recognitions: Science Advisor Hi, but that would mean that my calculation isn't right? About this "constant form": that should be zero, ofcourse; if a tensor is zero in one coordinate system, it's zero in every coordinate system. My bad ;)